United States presidential election in Oregon, 1988

United States presidential election in Oregon, 1988
Oregon
November 8, 1988
 
Nominee Michael Dukakis George H.W. Bush
Party Democratic Republican
Home state Massachusetts Texas
Running mate Lloyd Bentsen Dan Quayle
Electoral vote 7 0
Popular vote 616,206 560,126
Percentage 51.28% 46.61%

County Results
  Dukakis—60-70%
  Dukakis—50-60%
  Dukakis—<50%
  Bush—<50%
  Bush—50-60%
  Bush—60-70%
President before election

Ronald Reagan
Republican

 Elected President

George H. W. Bush
Republican

The 1988 United States presidential election in Oregon took place on November 8, 1988 as part of the 1988 United States presidential election. Voters chose seven electors of the Electoral College, who voted for president and vice president.

Oregon was won by the Democratic nominee, Massachusetts governor Michael Dukakis, over the Republican nominee, Vice President George H. W. Bush. Oregon was one of just 10 states won by Dukakis in an election won by Vice President Bush. It also marked the first victory by a Democratic presidential candidate in Oregon since 1964; Democrats have won every presidential election in Oregon since, albeit it nearly went red in 2000.[1]

Results

United States presidential election in Oregon, 1988[2]
Party Candidate Votes Percentage Electoral votes
Democratic Michael Dukakis 616,206 51.28% 7
Republican George H. W. Bush 560,126 46.61% 0
Libertarian Ron Paul 14,811 1.23% 0
New Alliance Party Lenora Fulani 6,487 0.54% 0
No party Write-in 3,974 0.33% 0
No party David Duke (write-in) 90 0.01% 0

References

  1. ↑ "Oregon Presidential Election Voting History". 270ToWin.com. Retrieved June 12, 2012.
  2. ↑ "1988 Presidential General Election Results - Oregon". Dave Leip's U.S. Election Atlas. Retrieved June 12, 2012.
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